1952 AMC 12 Problema 34

Question

El precio de un artículo aumentó un

p\%.

Después, el nuevo precio disminuyó un

p\%.

Si el precio final era de un dólar, el precio original era:

The price of an article was increased

p\%.

Later the new price was decreased

p\%.

If the last price was one dollar, the original price was:

\dfrac{1-p^2}{200}

\dfrac{\sqrt{1-p^2}}{100}

Un dólar

One dollar

1-\dfrac{p^2}{10{,}000-p^2}

\dfrac{10{,}000}{10{,}000-p^2}

Representa el aumento y la disminución mediante factores multiplicativos

Represent the increase and decrease by multiplicative factors

Si el precio original es

P,

entonces

P(1+\frac{p}{100})(1-\frac{p}{100})=1

If the original price is

P,

then

P(1+\frac{p}{100})(1-\frac{p}{100})=1

Si el precio original es de

P

dólares, después de ambos cambios el precio es

\begin{aligned} &P\left(1+\frac p{100}\right) \left(1-\frac p{100}\right)\\ &\qquad=P\left(1-\frac{p^2}{10{,}000}\right). \end{aligned}

Al igualar esto a

1

se obtiene

P=\frac{10{,}000}{10{,}000-p^2}.

Por lo tanto, la respuesta correcta es E.

If the original price is

P

dollars, then after both changes its price is

\begin{aligned} &P\left(1+\frac p{100}\right) \left(1-\frac p{100}\right)\\ &\qquad=P\left(1-\frac{p^2}{10{,}000}\right). \end{aligned}

Setting this equal to

1

gives

P=\frac{10{,}000}{10{,}000-p^2}.

Thus, the correct answer is E.

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